奇异行为:函数的三元运算符

Here is a simplified example of my problem. I thought that these functions would have exactly the same behavior:

这是我的问题的一个简化例子。我认为这些函数具有完全相同的行为:

def f1(l):
    if type(l[0][0])==list: f=lambda x:x[0][0]
    else: f=lambda x:x[0]
    l.sort(key=f,reverse=True)

def f2(l):
    f=lambda x:x[0][0] if type(l[0][0])==list else lambda x:x[0]
    l.sort(key=f,reverse=True)

l=[[1,2],[3,4]]

But in reality f1(l) works fine when f2(l) collapses with the exception:

但在现实中f1(l)在f2(l)崩溃时可以正常工作:

IndexError: list index out of range

So the question is why is it so and is it possible to use ternary operator that returns one of the functions at all?

问题是为什么会这样，是否有可能使用三元运算符返回一个函数?

1 个解决方案

#1

lambda has the lowest precedence among operators. This is why Python parses that line as

lambda在操作符中具有最低的优先级。这就是为什么Python将这一行解析为as

f = lambda x: (x[0][0] if type(l[0][0]) == list else lambda x: x[0])

The fix is to wrap individual lambdas in parentheses:

修正方法是将单个的lambdas括在括号中:

f = (lambda x: x[0][0]) if type(l[0][0]) == list else (lambda x: x[0])

That said, type(l[0][0]) == list is kinda wrong, isinstance(l[0][0], list) would be the best way (it also handles subclasses of list).

也就是说，type(l[0][0]) = list有点不对，isinstance(l[0][0]， list)是最好的方法(它也处理list的子类)。

#1